Question

A steel wire having cross-sectional area $1.5m{m}^2$ when stretched by a load produces a lateral strain $1.5\times {10}^{-5}$. Calculate the mass attached to the wire. $(Y_{steel}=2\times {10}^{11}N/m^2,Poisso{n}^{\prime }sratioa=0.291,g=9.8m/s^2)$

Answer 1

Given that $A$$=1.5m{m}^2$, lateral strain $=1.5\times {10}^{-5}$

$(Y_{steel}=2\times {10}^{11}N/m^2,Poisso{n}^{\prime }sratio\sigma =0.291,g=9.8m/s^2)$

$\sigma =\frac{Lateralstrain}{Logitudinalstrain}$

$0.291=\frac{1.5\times {10}^{-5}}{Logitudinalstrain}$

$\therefore$ Logitudinal strain$=\frac{1.5\times {10}^{-5}}{0.291}=5.14\times {10}^{-5}$

Logitudinal strain $=Y\times$ Logitudinal strain

$=2\times {10}^{11}\times 5.14\times {10}^{-5}$

$\therefore$ Logitudinal strain$=10.28\times {10}^6$

But Logitudinal strain $=\frac{Mg}{A}$

$\Rightarrow 10.28\times {10}^6=\frac{M\times 9.8}{1.5\times {10}^{-6}}$

$\Rightarrow M=\frac{10.28\times {10}^6\times 1.5\times {10}^{-6}}{9.8}=\frac{15.42}{9.8}$

$\therefore M=1.58kg$.